### Author Topic: FE-P5  (Read 1182 times)

#### Victor Ivrii

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##### FE-P5
« on: December 18, 2018, 06:18:35 AM »
Consider $P(z)= z^3 +2z -3-i$ and, using the argument theorem and Rouché's theorem calculate the number of its roots in each of the following domains:

(a)  $\{z\colon |z-1|<1\}$;

(b)  $\{z\colon |z-1|>1, |z|<2\}$,

(c) $\{z\colon |z|>2\}$.
« Last Edit: December 20, 2018, 05:06:11 AM by Victor Ivrii »

#### hz12

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##### Re: FE-P5
« Reply #1 on: December 18, 2018, 10:48:54 AM »
P(z)=z^3 -3 + 2z -i = (z^3 – 3)+(2z- i)

a, |z-1|<1
0 < z < 2
Using Rouche Theorem, let z = 2, z^3-3= 5 > 2z - i, z^3 -3 is the  dominant,
z^3-3 =0 and  z has 3 roots.
So it means when z<2, P(z) has 3 roots.
let z = 0, z^3-3 = -3 < 2z - i, 2z - i is the  dominant,
2z - i =0 and  z has 1 root.
So it means when z<0, f(z) has 1 root.

So when 0 < z < 2, f(z) has 3-1 =2 roots.

b, |z-1|>1, |z|<2
z < 0, z > 2, -2 < z < 2 so -2 < z < 0
when -2 < z < 0, it has 1 root.

c,  z > 2
when z < 2, it has 3 roots already, so when z > 2, it has 0 root.
« Last Edit: December 18, 2018, 10:50:52 AM by Hanyu Zhou »

#### yujiaha1

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##### Re: FE-P5
« Reply #2 on: December 18, 2018, 10:50:25 AM »

#### Ziqi Zhang

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##### Re: FE-P5
« Reply #3 on: December 18, 2018, 11:06:00 AM »
(a) P(z)=z3+2z−3−i=(z-1)3+3(z-1)2+5(z-1)-i
at |z−1|=1, |5(z−1)|≥|(z-1)3+3(z-1)2-i|
so at |z−1|<1, |5(z−1)| and P(z) have the same number of roots which is 1 (z=1).

(b) At |z|=2, |z3|≥|2z-3-i|
so at |z|<2, |z3| and P(z) have the same number of roots which is 3 (z=0 has the multiplicity of 3).
but we know |z−1|<1 is in |z|<2, z≠1, so at |z−1|>1,|z|<2, P(z) has the number of roots of 3-1=2.

(c) Because P(z) has the highest power of 3, so it has in all 3 roots, but we already find 3 roots at |z|<2,
so at |z|>2, the number of roots is 0.

#### Victor Ivrii

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##### Re: FE-P5
« Reply #4 on: December 20, 2018, 05:09:56 AM »
(a) As $|z|=2$ consider $Q(z)=z^3$; then  $|Q(z)|=8$,
$$|P(z)-Q(z)|=|2iz -3-i|\le 4+|3+i|=4+\sqrt{10}<8;$$
therefore $P$ has as many roots in $\{z\colon |z|<2\}$ as $Q(z)$ has, which is $3$ (we count orders).

(c) Then there are no roots in $\{z\colon |z|>2\}$.

(b) Consider $z=w+1$ with $|w|=1$; then $P(w+1)=w^3+3w^2 +5w -i$ and we set $Q(w)=5w$; then $|Q(w)|=5$,
$$|P(w)-Q(w)|=|w^3+3w^2 -i| < 5.$$
Indeed, $|P(w)-Q(w)|\le |w+3|+1< 5$ except $w=1$, and for $w=1$ we have $|w^3+3w^2 -i|=|4-i|=\sqrt{17}<5$. Therefore $P(w+1)$ has as many roots in $\{w\colon |w|<1\}$ as $Q(w)$ has, which is $1$ (we count orders).

Finally, in the domain  $\{z\colon |z-1|>1, |z|<2\}$, there are $3-1=2$ roots.